A car comes to a bridge during a storm and finds the bridge washed out. The driv

Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 18.4m above the river, whereas the opposite side is a mere 2.0mabove the river. The river itself is a raging torrent 58.0m wide.

How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?

What is the speed of the car just before it lands safely on the other side?

Explanation / Answer

The car will fall a vertical distance of 18.4 - 2 = 16.4 m vertically if it just clears the gap.

It has not initial y velocity, so we can apply d = vot + .5at2 to find how long it will be in the air

16.4 = (0) + (.5)(9.8)(t2)

t = 2.58 sec

That is the time the car will need to be in flight, so horizontally

d = vt

58 = (v)(2.58)

v = 22.48 m/s

b

The x velocity will not change, but the y velocity will. We can find it y speed as it hits the other side

vf2 = vo2 + 2ad

vf2 = 0 + (2)(9.8)(16.4)

vf = 17.92 m/s

Combine the x and y velocities by the pythagorean theorem

v = ?[(22.48)2 + (17.9)2]

v = 28.75 m/s

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